Question

Sodium phosphate, \(\mathrm{Na}_{3} \mathrm{PO}_{4},\) is made commercially by first neutralizing phosphoric acid with sodium carbonate to obtain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\). The \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) is further neutralized to \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) with \(\mathrm{NaOH}\) (a) Write net ionic equations for these reactions. (b) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is a much cheaper base than is \(\mathrm{NaOH}\) Why do you suppose that \(\mathrm{NaOH}\) must be used as well as \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to produce \(\mathrm{Na}_{3} \mathrm{PO}_{4} ?\)

Short answer

The net ionic reactions are:(a) \(\mathrm{H}_{3}\mathrm{PO}_{4} + \mathrm{Na}_{2}\mathrm{CO}_{3} \rightarrow 2\mathrm{Na}_{2}\mathrm{HPO}_{4} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\) (b) \(\mathrm{Na}_{2}\mathrm{HPO}_{4} + \(\mathrm{NaOH} \rightarrow 2\mathrm{Na}_{3} \mathrm{PO}_{4} + \mathrm{H}_{2}\mathrm{O}\). The use of both \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaOH}\) in the production of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) is necessary as \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is only capable of providing two \(Na^{+}\) ions and the third \(Na^{+}\) is provided by \(\mathrm{NaOH}\).

Step-by-step solution

Writing net ionic equation for \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) production

The reaction of phosphoric acid with sodium carbonate can be represented as follows: \(\mathrm{H}_{3}\mathrm{PO}_{4} + \mathrm{Na}_{2}\mathrm{CO}_{3} \rightarrow \mathrm{Na}_2\mathrm{HPO}_{4} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\). When this reaction equation is balanced, it becomes: \(\mathrm{H}_{3}\mathrm{PO}_{4} + \mathrm{Na}_{2}\mathrm{CO}_{3} \rightarrow 2\mathrm{Na}_{2}\mathrm{HPO}_{4} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\). This equation excludes spectator ions hence, this is a net ionic equation.

Writing net ionic equation for \(\mathrm{Na}_{2}\mathrm{HPO}_{4}\) conversion to \(\mathrm{Na}_{3} \mathrm{PO}_{4}\)

This reaction neutralize \(\mathrm{Na}_{2}\mathrm{HPO}_{4}\) to \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) with \(\mathrm{NaOH}\) and represented as \(\mathrm{Na}_{2}\mathrm{HPO}_{4} + \(\mathrm{NaOH} \rightarrow \mathrm{Na}_{3} \mathrm{PO}_{4} + \mathrm{H}_{2}\mathrm{O}\). After balancing this equation it becomes: \(\mathrm{Na}_{2}\mathrm{HPO}_{4} + \(\mathrm{NaOH} \rightarrow 2\mathrm{Na}_{3} \mathrm{PO}_{4} + \mathrm{H}_{2}\mathrm{O}\). Spectator ions are already excluded from this equation, thus this also represents the net ionic equation.

Explanation of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaOH}\) usage

\(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is only capable of providing two \(Na^{+}\) ions while the manufacturing of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) requires three \(Na^{+}\) ions. Therefore, another base with \(Na^{+}\) ion is needed. So, \(\mathrm{NaOH}\) is used in this reaction as it provides an extra \(Na^{+}\) needed to fully neutralize phosphate ion \(\mathrm{(PO_{4}^{-3})}\) to produce \(\mathrm{Na}_{3} \mathrm{PO}_{4}\).

Key concepts

Sodium Phosphate Production

Sodium phosphate, known chemically as \( \mathrm{Na}_3 \mathrm{PO}_4 \), is produced through a multi-step process involving the neutralization of phosphoric acid. The process begins with phosphoric acid \( (\mathrm{H}_3\mathrm{PO}_4) \) reacting with sodium carbonate \( (\mathrm{Na}_2\mathrm{CO}_3) \) to form an intermediate product, disodium hydrogen phosphate \( (\mathrm{Na}_2\mathrm{HPO}_4) \). This occurs because sodium carbonate is capable of replacing only two of the hydrogen ions in phosphoric acid with sodium ions, producing carbon dioxide \( (\mathrm{CO}_2) \) and water \( (\mathrm{H}_2\mathrm{O}) \) as byproducts.
The first net ionic equation from this neutralization is:\[ \mathrm{H}_3\mathrm{PO}_4 + \mathrm{Na}_2\mathrm{CO}_3 \rightarrow \mathrm{Na}_2\mathrm{HPO}_4 + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}. \]
It's essential to recognize the roles of the substances used in this step. Phosphoric acid acts as a triprotic acid, losing two protons while leaving one proton attached to the phosphate ion. Sodium carbonate, being a basic salt, increases the basicity required to partially neutralize the acid.The process then involves a subsequent neutralization reaction where \( \mathrm{Na}_2\mathrm{HPO}_4 \) is further treated with sodium hydroxide \( (\mathrm{NaOH}) \) to convert it fully into sodium phosphate. It's crucial for the process to achieve a complete reaction, yielding a fully neutralized phosphate ion. The net ionic equation for this step is:\[ \mathrm{Na}_2\mathrm{HPO}_4 + \mathrm{NaOH} \rightarrow \mathrm{Na}_3\mathrm{PO}_4 + \mathrm{H}_2\mathrm{O}. \]
By understanding these steps in the production of sodium phosphate, we observe the importance of each base used in achieving the desired chemical compound.

Neutralization Reactions

Neutralization is a fundamental chemical reaction, occurring when an acid and a base react to form a salt and water. In the context of sodium phosphate production, the neutralization process involves initially reacting phosphoric acid with a base to form a partially neutralized salt. This is followed by another reaction to fully neutralize the acid.
There are key characteristics of neutralization reactions that make them noteworthy:- The reaction of an acid with a base typically produces water as one of the products.- An ionic compound, often referred to as a 'salt', is formed as a product of neutralization, which in this context is sodium phosphate.- Neutralization reactions often involve proton transfer, where the acid donates protons \((H^+)\) and the base supplies hydroxide ions \((OH^-)\) or ionic equivalents.
In sodium phosphate's production, the sequence of neutralization ensures the transformation from \( \mathrm{H}_3\mathrm{PO}_4 \) to \( \mathrm{Na}_3\mathrm{PO}_4 \) through intermediate stages involving base addition. Each reaction stage is essential, ensuring effective proton replacement and achieving a stable ionic structure in the final product.

Base Selection in Synthesis

Choosing suitable bases in chemical synthesis is critical for successful compound formation. In sodium phosphate production, the choice to initially use sodium carbonate \( (\mathrm{Na}_2\mathrm{CO}_3) \) is economical due to its lower cost. Sodium carbonate effectively replaces two of the hydrogen ions from phosphoric acid, making it an initial target base. Its capability in neutralizing acids while being cost-effective makes it strategically valuable for large-scale production.
However, the transformation to sodium phosphate requires replacing the final hydrogen with sodium, necessitating the use of sodium hydroxide \( (\mathrm{NaOH}) \). This step is crucial for providing the additional sodium ion required for complete neutralization to \( \mathrm{Na}_3\mathrm{PO}_4 \). Despite the higher cost, \( \mathrm{NaOH} \)'s role cannot be sidelined because:

• It introduces an extra sodium ion that sodium carbonate cannot provide.
• It fully neutralizes the phosphate ion \((\mathrm{PO}_4^{3-})\) ensuring the formation of the correct compound.
• Sodium hydroxide’s strong basicity facilitates the complete neutralization reaction, ensuring a high yield of the desired product.

Understanding why specific bases are chosen in synthesis helps us appreciate the nuances of chemical reactions and the economic considerations involved in industrial practices.